/*
 * Scramble String
Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = "great":

    great
   /    \
  gr    eat
 / \    /  \
g   r  e   at
           / \
          a   t
To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".

    rgeat
   /    \
  rg    eat
 / \    /  \
r   g  e   at
           / \
          a   t
We say that "rgeat" is a scrambled string of "great".

Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".

    rgtae
   /    \
  rg    tae
 / \    /  \
r   g  ta  e
       / \
      t   a
We say that "rgtae" is a scrambled string of "great".

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.
 */
package com.xinpan.exercise;

public class IsScramble {
    public boolean isScramble(String s1, String s2) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int l1 = s1.length();
        int l2 = s2.length();
        if(l1 != l2)
            return false;
        else if (l1 == 1)
            return s1.charAt(0) == s2.charAt(0);
        else if(l1 == 2)
        {
            if(s1.charAt(0) == s2.charAt(0) && s1.charAt(1) == s2.charAt(1) ||
                s1.charAt(0) == s2.charAt(1) && s1.charAt(1) == s2.charAt(0))
                return true;
            else
                return false;
        }
        
        if(!sameCharSet(s1, s2))
            return false;
            
        char val = s1.charAt(0);
        int i = 0;
        for(; i < l1/2; i++)
            if(s2.charAt(i) == val || s2.charAt(l1-i-1) == val)
                break;

        for(i++; i < l1; i++)
        {
            if(isScramble(s1.substring(0, i), s2.substring(0, i)) &&
                isScramble(s1.substring(i, l1), s2.substring(i, l1)))
                return true;
            if(isScramble(s1.substring(0, i), s2.substring(l1-i, l1)) &&
                isScramble(s1.substring(i, l1), s2.substring(0, l1-i)))
                return true;
        }

            
        return false;
    }
    
    public boolean sameCharSet(String s1, String s2)
    {
        int[] c = new int[256];
        for(int i = 0; i < 256; i++)
            c[i] = 0;
            
        for(int i = 0; i < s1.length(); i++)
        {
            c[(int)s1.charAt(i)]++;
            c[(int)s2.charAt(i)]--;
        }
        for(int i = 0; i < 256; i++)
            if(c[i] != 0)
                return false;
        return true;
    }
    
    public static void main(String[] args)
    {
    	IsScramble is = new IsScramble();
    	is.isScramble("abcdd", "dbdac");
    }
}
